Question

Find equation of circle which touches the x-axis and whose centre is (3,4)

Answer 1

C =(3,4)

x-axis is the tangent to the circle.
so, centre is on a line normal to the x-axis.
point of tangent = (3,0)

radius = 4-0 = 4

eqn of the circle:

${(x - 3)}^{2} + {(y - 4)}^{2} = {4}^{2} \\ = > {x}^{2} + {y}^{2} - 6x - 8y + 9 = 0$

Answer 2

Answer:

The equation of circle which touches the x-axis and with centre (3,4) is  x²+y² − 6x − 8y + 9 = 0.

Step-by-step explanation:

The equation of the circle with centre (h,k) and radius r is given by (x-h)² + (y-k)² = r².

Given the circle touches the x-axis, then the y-coordinate of its center will be the radius of the circle.

Thus its center is (3,4) and the radius is 4.

So its equation  (x−3)²+(y−4)²=4²

⇒ (x²+3²-2*x*3) + (y²+4²-2*y*4) = 16

⇒ x² + 9 -6x + y² + 16 - 8y = 16

⇒ x² + y² - 6x - 8y = 16 - 16 - 9

⇒ x² + y² - 6x - 8y = -9

⇒ x²+y² − 6x − 8y + 9 = 0

therefore the equation of the circle is x²+y² − 6x − 8y + 9 = 0.